# Why is this Brownian motion event not measurable w.r.t. the natural filtration?

In my lecture on stochastic processes it is stated that the natural filtration $$\mathcal{F_t}^0=\sigma(\forall s\leq t: \omega\mapsto \omega(s)$$ is measurable$$)$$ is not a good choice for Brownian motion because for example $$L=\lim_{t\searrow 0}\frac{B_t}{\sqrt{2t\log|\log t|}}$$ is not measurable w.r.t. $$\mathcal{F}^0$$ but w.r.t. $$\mathcal{F_t}=\bigcap_{s>t}\mathcal{F_s}^0$$.

Why is this the case? I have no clue why $$L$$ is $$\mathcal{F}$$-measurable and why not $$\mathcal{F}^0$$-measurable and it's nowhere explained.

Can someone of you help me please?

Thanks

• The second filtration is right continuous – badatmath Jan 7 at 16:00
• The natural filtration is such that $\mathcal{F_0^0}=\{\phi, \Omega\}$ because it's trivial sigma algebra (a $B_0=0$), so if $L$ is not constant you are done. Using time inversion you can look at the law of iterated logarithm to get a non constant result for L unless mistaken. – TheBridge Jan 7 at 16:56

Notice that for each $$\epsilon>0$$, $$L=\inf_{0< s<\epsilon, s\in\Bbb Q}\sup_{t\in(0,s), t\in\Bbb Q}{B_t\over \sqrt{2t\log|\log t|}}$$ is $$\mathcal F^0_\epsilon$$ measurable. (I have replaced the limit in your definition of $$L$$ with limit superior; the latter always exists, the former doesn't exist, for a.e. Brownian path.) It follows that $$L$$ is $$\mathcal F_0$$ measurable. On the other hand, $$\mathcal F^0_0=\sigma\{B_0\}$$ and $$L$$ is clearly not a function of $$B_0$$ alone.